Try this beautiful problem from AMC 10A, 2007 based on Numbers on cube.
The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?
Number system
adition
Cube
Answer: \(18\)
AMC-10A (2007) Problem 11
Pre College Mathematics
Given condition is "The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same".so we may say that if we think there is a number on the vertex then it will be counted in different faces also.
can you finish the problem........
Therefore we have to count the numbers \(3\) times so the total sum will be \(3(1+2+....+8)\)=\(108\)
can you finish the problem........
Now there are \(6\) faces in a Cube.....so the common sum will be \(\frac{108}{6}\)=\(18\)
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